You are given a 2D array queries, where queries[i] is of the form [l, r]. Each queries[i] defines an array of integers nums consisting of elements ranging from l to r, both inclusive.
In one operation, you can:
\n\na and b from the array.floor(a / 4) and floor(b / 4).Your task is to determine the minimum number of operations required to reduce all elements of the array to zero for each query. Return the sum of the results for all queries.
\n\n\n
Example 1:
\n\nInput: queries = [[1,2],[2,4]]
\n\nOutput: 3
\n\nExplanation:
\n\nFor queries[0]:
nums = [1, 2].nums[0] and nums[1]. The array becomes [0, 0].For queries[1]:
nums = [2, 3, 4].nums[0] and nums[2]. The array becomes [0, 3, 1].nums[1] and nums[2]. The array becomes [0, 0, 0].The output is 1 + 2 = 3.
Example 2:
\n\nInput: queries = [[2,6]]
\n\nOutput: 4
\n\nExplanation:
\n\nFor queries[0]:
nums = [2, 3, 4, 5, 6].nums[0] and nums[3]. The array becomes [0, 3, 4, 1, 6].nums[2] and nums[4]. The array becomes [0, 3, 1, 1, 1].nums[1] and nums[2]. The array becomes [0, 0, 0, 1, 1].nums[3] and nums[4]. The array becomes [0, 0, 0, 0, 0].The output is 4.
\n\n
Constraints:
\n\n1 <= queries.length <= 105queries[i].length == 2queries[i] == [l, r]1 <= l < r <= 109给你一个二维数组 queries，其中 queries[i] 形式为 [l, r]。每个 queries[i] 表示了一个元素范围从 l 到 r （包括 l 和 r ）的整数数组 nums 。
在一次操作中，你可以：
\n\na 和 b。floor(a / 4) 和 floor(b / 4)。你的任务是确定对于每个查询，将数组中的所有元素都变为零的 最少 操作次数。返回所有查询结果的总和。
\n\n\n\n
示例 1：
\n\n输入： queries = [[1,2],[2,4]]
\n\n输出： 3
\n\n解释：
\n\n对于 queries[0]：
nums = [1, 2]。nums[0] 和 nums[1]。数组变为 [0, 0]。对于 queries[1]：
nums = [2, 3, 4]。nums[0] 和 nums[2]。数组变为 [0, 3, 1]。nums[1] 和 nums[2]。数组变为 [0, 0, 0]。输出为 1 + 2 = 3。
示例 2：
\n\n输入： queries = [[2,6]]
\n\n输出： 4
\n\n解释：
\n\n对于 queries[0]：
nums = [2, 3, 4, 5, 6]。nums[0] 和 nums[3]。数组变为 [0, 3, 4, 1, 6]。nums[2] 和 nums[4]。数组变为 [0, 3, 1, 1, 1]。nums[1] 和 nums[2]。数组变为 [0, 0, 0, 1, 1]。nums[3] 和 nums[4]。数组变为 [0, 0, 0, 0, 0]。输出为 4。
\n\n\n
提示：
\n\n1 <= queries.length <= 105queries[i].length == 2queries[i] == [l, r]1 <= l < r <= 109x, the number of \"/4\" operations to change it to 0 is floor(log4(x)) + 1.",
"Always pair the 2 numbers with the maximum \"/4\" operations needed."
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